Problem 60 - Calculus and its applications

Question

Is the function given by $$ F(x)=\left\\{\begin{array}{ll} \frac{x^{2}-1}{x-1}, & \text { for } x \neq 1, \\ 4, & \text { for } x=1 \end{array}\right. $$ continuous at \(x=1 ?\) Why or why not?

Step-by-Step Solution

Answer

The function is not continuous at \(x=1\) because the limit and the function value differ.

Step 1: Understand Continuity at a Point

A function is continuous at a point if three conditions are met: the function is defined at that point, the limit of the function exists as it approaches the point, and the limit equals the function's value at that point. We will apply these criteria to check continuity at \(x=1\).

Step 2: Value of the Function at x=1

For \(x=1\), the function \(F(x)\) is explicitly defined as 4. Thus, \(F(1) = 4\). This satisfies the first requirement for continuity.

Step 3: Find the Limit As x Approaches 1

Since the function expression \(\frac{x^2 - 1}{x - 1}\) applies for \(x eq 1\), we simplify it to find the limit as \(x\) approaches 1. The expression simplifies using factorization: \[\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \text{ (for } x eq 1\text{)}\]. Therefore, \(\lim_{x \to 1} F(x) = 1 + 1 = 2\).

Step 4: Compare Limit to Function Value

We now compare the limit as \(x\) approaches 1 to the value of the function at \(x=1\): \(\lim_{x \to 1} F(x) = 2\) and \(F(1) = 4\). Since \(2 eq 4\), the third condition for continuity is not met.

Key Concepts

Continuity

In calculus, continuity is a fundamental concept that ensures a function holds no breaks, jumps, or holes at a point. For a function to be continuous at a point, three criteria must be satisfied:

• The function must be defined at that point, meaning there is an actual, finite value.
• The limit of the function must exist as it approaches the point from both sides.
• The calculated limit must equal the function's value at that point.

When these three conditions are met, you can smoothly trace the function's path without lifting your pencil. In the given exercise, the piecewise function is evaluated for continuity at \(x=1\). The fact that the limit and function value at this point differ indicates discontinuity.

Limits

Limits are a foundational aspect of calculus, representing the behavior of a function as the input approaches a particular value.
Understanding limits is crucial for analyzing functions at points where they may not be explicitly defined. For example, sometimes you have undefined points or expressions like \(rac{0}{0}\).
For the piecewise function in the exercise, the continuity at \(x=1\) depends on whether the limit of the function as \(x\) approaches 1 matches the function's defined value there. We simplify the expression \(\frac{x^2 - 1}{x - 1}\) to \(x + 1\) using factorization. As \(x\) approaches 1, the limit is clearly 2. This differs from the function's value of 4 at \(x=1\), leading to a failure in meeting the continuity criteria.

Piecewise Functions

Piecewise functions are those defined by different expressions for different intervals of the domain. They are segmented based on condition criteria, creating varying behavior within a single function.
A common scenario involves using constraints on the input to define these segments. For instance, one expression might apply to all values except a particular point, while another might define the exact value at that point. Such is the case in the exercise, where \(\frac{x^2 - 1}{x - 1}\) applies for all \(x eq 1\), and a constant value of 4 is defined specifically at \(x=1\).
These functions help handle cases like undefined expressions by providing a direct value where the expression doesn't apply. Yet, this approach can introduce points of discontinuity if the limits don't match this value, as demonstrated in our exercise. Understanding how different parts of a piecewise function interact is vital for ensuring overall function continuity.